Let us apply the bernoulli principle between states 1

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Let us apply the Bernoulli principle between states 1 and 3, the pipe interior and the peak of the height of the fountain. We will estimate the velocity of the water in the pipe to be very small, v 1 0 m/s . We will also estimate the velocity at the apex of the motion to be negligible, v 3 = 0 m/s . Let us apply Eq. (9.44): P 1 ρ + 1 2 v 2 1 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright 0 + gz 1 = P 3 ρ + 1 2 v 2 3 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright =0 + gz 3 , (9.65) P 1 = P atm + ρg ( z 3 z 1 ) , (9.66) P 1 P atm bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright = P gauge = ρg ( z 3 z 1 ) . (9.67) Substituting numbers, we find P gauge = parenleftbigg 997 kg m 3 parenrightbigg parenleftBig 9 . 81 m s 2 parenrightBig (30 m ) = 2 . 95 × 10 5 Pa = 293 . 4 kPa. (9.68) CC BY-NC-ND. 2011, J. M. Powers.
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276 CHAPTER 9. SECOND LAW ANALYSIS FOR A CONTROL VOLUME ρ = 997 kg/m 3 , P 1 = ?, v 1 ~ 0 m/s v 3 ~ 0 m/s, P 3 = 100 kPa v 2 = ?, P 2 = 100 kPa 30 m g = 9.81 m/s 2 Figure 9.8: Sketch of simple water fountain. We can use the same principle to estimate the exit velocity, v 2 . Here we take z 1 z 2 . P 1 ρ + 1 2 v 2 1 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright 0 + gz 1 = P 2 ρ + 1 2 v 2 2 + g z 2 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright z 1 , (9.69) P 1 ρ = P atm ρ + 1 2 v 2 2 , (9.70) v 2 = radicalBigg 2( P 1 P atm ) ρ , (9.71) v 2 = radicalBigg 2 P gauge ρ . (9.72) Substituting numbers, we find v 2 = radicalBigg 2(2 . 934 × 10 5 Pa ) 997 kg m 3 = 24 . 26 m s . (9.73) One could use a similar analysis to estimate the necessary pressure to generate the jet of the University of Notre Dame’s War Memorial Fountain, depicted in Fig. 9.9. Example 9.4 Perform a similar calculation for the problem sketched in Fig. 9.8, but account for mass conservation. Take the cross-sectional area of the pipe to be A 1 = A 4 = 1 m 2 , and that of the hole to be A 2 = 0 . 01 m 2 . We measure v 1 = 1 m/s . See Fig. 9.10. Assume we have the same P gauge = 293 . 4 kPa as calculated earlier. Find the new height of the fountain, z 3 , and the new exit velocity v 2 . CC BY-NC-ND. 2011, J. M. Powers.
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9.2. BERNOULLI’S PRINCIPLE 277 Figure 9.9: The University of Notre Dame’s War Memorial Fountain, 4 June 2010. m 1 . m 2 . m 4 . A 2 = 0.01 m 2 A 1 = 1 m 2 A 4 = 1 m 2 v 1 = 1 m/s Figure 9.10: Schematic of mass balance for water fountain problem. The mass balance gives us dm cv dt bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright =0 = ˙ m 1 ˙ m 2 ˙ m 4 , (9.74) 0 = ˙ m 1 ˙ m 2 ˙ m 4 , (9.75) ˙ m 4 = ˙ m 1 ˙ m 2 . (9.76) We recall that ˙ m = ρ v A , so ρ 4 v 4 A 4 = ρ 1 v 1 A 1 ρ 2 v 2 A 2 . (9.77) We assume incompressible flow, so ρ 1 = ρ 2 = ρ 4 = ρ , and we have A 1 = A 4 , so ρ v 4 A 1 = ρ v 1 A 1 ρ v 2 A 2 , (9.78) v 4 = v 1 v 2 A 1 A 2 . (9.79) This is nice, but not that useful. It simply predicts a lessening of velocity downstream of the hole. CC BY-NC-ND. 2011, J. M. Powers.
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278 CHAPTER 9. SECOND LAW ANALYSIS FOR A CONTROL VOLUME Bernoulli’s equation applied between 1 and 2 gives P 1 ρ + 1 2 v 2 1 + gz 1 = P 2 ρ + 1 2 v 2 2 + g z 2 bracehtipupleftbracehtipdownrightbracehtipdownleftbracehtipupright z 1 , (9.80) P 1 ρ + 1 2 v 2 1 = P atm ρ + 1 2 v 2 2 , (9.81) v 2 = radicalBigg 2( P 1 P atm ) ρ + 1 2 v 2 1 , (9.82) = radicalBigg 2 P gauge ρ + 1 2 v 2 1 . (9.83) With numbers, we get v 2 = radicalBigg 2(2 . 934
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